Ch.7 Circuits

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Overview

• DC Circuits, potentials and currents are constant
• analyzed using Kirchhoff's junction law (charge conservation) and loop law (energy conservation)
• Ohm's law used only for resistors
• power: rate of battery supplying energy/rate of resistor dissipating energy (watts)
• resistors can be combined like capacitors to an equivalent resistance in series or parallal
• RC circuits charge/discharge capacitors through resistors exponentially with time constant $\tau=RC$

Circuit Diagram

used to describe/analyze circuits

• battery (+ is long end, - is short)
• wave-y thing is resistor
• line is wire
• $-||-$ shape is capacitor

Analyzing Circuits

We want to find potential difference across each circuit component and current in each component

From conservation of charge,
$\sum I_\text{in}=\sum I_\text{out}$
From conservation of energy, a charge that moves around a closed path has $\Delta U_\text{elec}=0$, so the potential differences of components satisfy
$\Delta V_\text{loop}=\sum(\Delta V)_i=0$
this is Kirchoff's Loop Law
For batteries, $\Delta V_\text{bat}=\varepsilon$ going from $-$ end to $+$ end
For resistors, $\Delta V_\text{res}=-\Delta V_R=-IR$; using Ohm's Law gives the magnitude $IR$

Basic Circuits

The most basic is a battery (a source) connected to a resistor, called a load (could be like a "$10\Omega$ resistor" or a lightbulb). A continuous path between the battery terminals form a complete circuit

Power

power is the rate at which energy is supplied (e.g. to a charge by a battery) measured in $\text{J}/\text{S}$ or $\text{W}$ (watts)
$P_\text{bat}=\frac{dU}{dt}=\frac{dq}{dt}\varepsilon=I\varepsilon$
Since all energy from the battery gets dissipated by the resistor, the resistor's power is
$P_R=\frac{dq}{dt}\Delta V_R=I\Delta V_R=I\varepsilon$
In a simple circuit with one battery and one resistor, the power and potential differences are equal.
Across a resistor, Ohm's Law can be used so
$P_R=I\Delta V_R=I^2R=\frac{(\Delta V_R)^2}{R}$

The total energy dissipated can be calculated as $J=P_R\Delta t$. If $P_R$ kilowatts of power is used over $\Delta t$ hours then the load has used $P_R\Delta t\text{ kWh}$ (kiloWatt hours) of energy.

An ammeter measures current; must be in series

Real Batteries

Real batters contain an internal resistance, symbolized $r$ so the terminal voltage across it is $\varepsilon-Ir$

If there was no load, the wire would short the circuit, with only internal load having resistance, giving a current of $\varepsilon/r$

A short circuit occurs when a wire that is normally separated by a high resistance is separated instead by a very low resistance. When $R=0$, the current is
$I_\text{short}=\frac{\varepsilon}{r}$
where $r$ is the internal resistance of the battery. This creates the maximum possible current

Parallel Resistors

Resistors where both ends are connected are parallel resistors; the connected ends have the same potential, giving the same potential differences for both. Thus, using the junction law,
$I=\sum I_i=\sum\frac{\Delta V}{R_i}=\Delta V\sum\frac{1}{R_i}$
The equivalent resistance is then therefore
$=\frac{I}{\Delta V}=\frac{1}{R_\text{eq}}=\sum\frac{1}{R_i}$

In summary, the current is the same in series and voltages are added; the voltage is the same in parallel and current is added

A voltmeter measures voltages by being in parallel with a circuit. The ideal voltmeter has resistance $R_\text{voltmeter}=\infty\space\Omega$

Grounded Circuits

Suppose we want to connect two different circuits. The potentials may not be compatible because we have only needed to consider the differences across components within that circuit.
Call the potential of the earth $V_\text{earth}=0\text{ V}$ and connect one wire to it. This creates an incomplete circuit, leading to an equipotential wire with $0$ current. Thus, this sets one part of the circuit to be $0\text{ V}$ relative to the earth. This circuit is not grounded by the ground wire

RC Circuits

A circuit with resistors and capacitors have a current that varies with time as the capacitor charges/discharges is called an RC Circuit

• good for timekeeping circuits

Consider a simple circuit with capacitor charged to $Q_0$ and potential difference $\Delta V_0=Q_0/C$, a resistor, and a switch. When the switch is open, there is no current, so $\Delta V_0=0$
As soon as the switch is closed at time $t=0$, the capacitor starts to discharge and creates a current.

What is the current at time $t$? How long does it take to discharge?
By the loop law,
$\Delta V_\text{cap}+\Delta V_\text{res}=\frac{Q}{C}-IR=0$
where $Q$ and $I$ are the instantaneous values at time $t$. Since we have $I=dq/dt$ as the rate of charge flow through the resistor, we have $I=-\frac{dQ}{dt}$
as flowing charge was charge removed from the capacitor. When $Q$ is decreasing, $dQ/dt$ is negative, so $I$ is positive, as expected. This turns the loop law equation into
$\frac{Q}{C}+\frac{dQ}{dt}R=0\implies-\frac{dt}{RC}=\frac{dQ}{Q}\implies\ln{Q}\Big|_{Q_0}^Q=-\frac{t}{RC}\Big|_0^t\implies Q=Q_0e^{-t/RC}$
Since the exponent must be dimensionless, the quantity $RC$ must have the dimension of time. We call the time constant$\tau=RC$ (note this means an Ohm-Farad must be a second)
With $\Delta V$ proportional to $Q$, this gives the equations
$Q=Q_0e^{-t/\tau},\space\space\space\Delta V=\Delta V_0e^{-t/\tau}$
This results in the current being
$I=-\frac{dQ}{dt}=\frac{Q_0}{\tau}e^{-t/\tau}=\frac{\Delta V_0}{R}e^{-t/\tau}=I_0e^{-t/\tau}$
where $I_0$ is the initial current the moment the switch is closed.

Charging a Capacitor

Consider a simple circuit with a battery with emf $\varepsilon$ and switch, a resistor, and capacitor. When the switch closes, the capacitor charges until $\Delta V_\text{cap}=\varepsilon$, at which point it has charge $Q_0=C\Delta V_\text{cap}=C\varepsilon$
Using the loop law,
$\Delta V_\text{bat}+\Delta V_\text{res}+\Delta V_\text{cap}=0\\\varepsilon-IR-\frac{Q}{C}=0$
This time, the current is the rate of the capacitor's charge increasing, so $I=dQ/dt$, giving
$\frac{\varepsilon}{R}-\frac{Q}{CR}=\frac{C\varepsilon-Q}{CR}=\frac{dQ}{dt}$
Finally, we solve the differential equation
$\int_0^t\frac{dt}{CR}=\int_0^Q\frac{dQ}{C\varepsilon-Q}\implies \frac{t}{CR}=-\ln(C\varepsilon-Q)+\ln(C\varepsilon)=-\ln\left(\frac{C\varepsilon-Q}{C\varepsilon}\right)$
With some algebra and substituting $\tau=CR$, we have
$C\varepsilon e^{-t/\tau}=C\varepsilon-Q\implies Q=C\varepsilon(1-e^{-t/\tau})$
and from $\Delta V=Q/C$, we have
$\Delta V=\varepsilon(1-e^{-t/\tau})$
The resulting current from $I=dQ/dt$ is
$I=\frac{C\varepsilon}{\tau}e^{-t/\tau}=\frac{C\varepsilon}{RC}e^{-t/\tau}=\frac{\varepsilon}{R}e^{-t/\tau}=I_0e^{-t/\tau}$
where $\tau=RC$ is the time constant and $I_0$ is the maximum current, which is when $t=0$.